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zaharov [31]
3 years ago
8

How do you solve 90% of 20?

Mathematics
1 answer:
Marizza181 [45]3 years ago
3 0
All you would do is multiply .9 by 20
.9*20=18

90% of 20 is 18
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If f(x)=5x-13, then f^-1(x)=
svet-max [94.6K]

Answer:

Hello,

If f(x)=5x-13, then f^-1(x)= \frac{1}{5x-13}

have good day


Step-by-step explanation:


8 0
3 years ago
find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM​
Snezhnost [94]

\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

  • Given - <u>A </u><u>trapezium</u><u> </u><u>ABCD </u><u>with </u><u>non </u><u>parallel </u><u>sides </u><u>of </u><u>measure </u><u>1</u><u>5</u><u> </u><u>cm </u><u>each </u><u>!</u><u> </u><u>along </u><u>,</u><u> </u><u>the </u><u>parallel </u><u>sides </u><u>are </u><u>of </u><u>measure </u><u>1</u><u>3</u><u> </u><u>cm </u><u>and </u><u>2</u><u>5</u><u> </u><u>cm</u>

  • To find - <u>Area </u><u>of </u><u>trapezium</u>

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

<u>Construction</u><u> </u><u>-</u>

draw \: CE \: \parallel \: AD \:  \\ and \: CD \: \perp \: AE

Now , we can clearly see that AECD is a parallelogram !

\therefore AE = CD = 13 cm

Now ,

AB = AE + BE \\\implies \: BE =AB -  AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm

Now , In ∆ BCE ,

semi \: perimeter \: (s) =  \frac{15 + 15 + 12}{2}  \\  \\ \implies \: s =  \frac{42}{2}  = 21 \: cm

Now , by Heron's formula

area \: of \: \triangle \: BCE =  \sqrt{s(s - a)(s - b)(s - c)}  \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)}  \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21}  \: cm {}^{2}

Also ,

area \: of \: \triangle \:  =  \frac{1}{2}  \times base \times height \\  \\\implies 18 \sqrt{21} =  \: \frac{1}{\cancel2}  \times \cancel12  \times height \\  \\ \implies \: 18 \sqrt{21}  = 6 \times height \\  \\ \implies \: height =  \frac{\cancel{18} \sqrt{21} }{ \cancel 6}  \\  \\ \implies \: height = 3 \sqrt{21}  \: cm {}^{2}

<u>Since </u><u>we've </u><u>obtained </u><u>the </u><u>height </u><u>now </u><u>,</u><u> </u><u>we </u><u>can </u><u>easily </u><u>find </u><u>out </u><u>the </u><u>area </u><u>of </u><u>trapezium </u><u>!</u>

Area \: of \: trapezium =  \frac{1}{2}  \times(sum \: of \:parallel \: sides) \times height \\  \\ \implies \:  \frac{1}{2}  \times (25 + 13) \times 3 \sqrt{21}  \\  \\ \implies \:  \frac{1}{\cancel2}  \times \cancel{38 }\times 3 \sqrt{21}  \\  \\ \implies \: 19 \times 3 \sqrt{21}  \: cm {}^{2}  \\  \\ \implies \: 57 \sqrt{21}  \: cm {}^{2}

hope helpful :D

6 0
2 years ago
I need help with my math homework x+23=9/10
BartSMP [9]

Answer:

-221/10

Step-by-step explanation:

In the attached file

8 0
3 years ago
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